Find $\lim_{x\to 4}\dfrac{2-\sqrt{4x-12}}{x-4}$. Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $1$ (Choice C) C $-1$ (Choice D) D The limit doesn't exist
Answer: Substituting $x=4$ into $\dfrac{2-\sqrt{4x-12}}{x-4}$ results in the indeterminate form $\dfrac{0}{0}$. This doesn't necessarily mean the limit doesn't exist, but it does mean we have to work a little before we find it. Since we have a rational expression with a square root on our hands, let's try to re-write it using the method of rationalization. $\begin{aligned} &\phantom{=}\dfrac{2-\sqrt{4x-12}}{x-4} \\\\ &=\dfrac{2-\sqrt{4x-12}}{x-4}\cdot\dfrac{2+\sqrt{4x-12}}{2+\sqrt{4x-12}} \gray{\text{Rationalize the numerator}} \\\\ &=\dfrac{2^2-(4x-12)}{(x-4)(2+\sqrt{4x-12})} \\\\ &=\dfrac{-4\cancel{(x-4)}}{\cancel{(x-4)}(2+\sqrt{4x-12})} \gray{\text{Cancel out common factors}} \\\\ &=\dfrac{-4}{2+\sqrt{4x-12}} \text{, for }x\neq 4 \end{aligned}$ This means that the two expressions have the same value for all $x$ -values (in their domains) except for $4$. We can now use the following theorem: If $f(x)=g(x)$ for all $x$ -values in a given interval except for $x=c$, then $\lim_{x\to c}f(x)=\lim_{x\to c}g(x)$. In our case, $\dfrac{2-\sqrt{4x-12}}{x-4}=\dfrac{-4}{2+\sqrt{4x-12}}$ for all $x$ -values in the interval $(3.5,4.5)$ except for $x=4$. Therefore, $\lim_{x\to 4}\dfrac{2-\sqrt{4x-12}}{x-4}=\lim_{x\to 4}\dfrac{-4}{2+\sqrt{4x-12}}=-1$. (The last limit was found using direct substitution.) [I want to see how this looks graphically!] In conclusion, $\lim_{x\to 4}\dfrac{2-\sqrt{4x-12}}{x-4}=-1$.